Math Problem Statement

The mass of a bag of rice, in kg, is normally distributed with mean μ and standard deviation σ. If 97.5% of the bags of rice have masses of more than 30.5 kg and 88.49% of the bags of rice have masses of less than 46.3 kg, find the value of μ and σ.

Solution

The problem presented involves a normal distribution of rice bag weights. Here's a breakdown:

  • The weights are normally distributed with mean μ\mu and standard deviation σ\sigma.
  • 97.5% of the bags weigh more than 30.5 kg.
  • 88.49% of the bags weigh less than 46.3 kg.
  • We are tasked with finding the values of μ\mu and σ\sigma.

Step 1: Use of Z-scores

From the standard normal distribution table (or a calculator), we can convert the given percentages into Z-scores:

  1. 97.5% of the bags weigh more than 30.5 kg:

    • This means 2.5% weigh less than 30.5 kg.
    • From the Z-table, the Z-score corresponding to 2.5% in the lower tail is approximately -1.96.
  2. 88.49% of the bags weigh less than 46.3 kg:

    • From the Z-table, the Z-score corresponding to 88.49% is approximately 1.19.

Step 2: Set up the equations

Using the Z-score formula:
Z=XμσZ = \frac{X - \mu}{\sigma} where XX is the data value, μ\mu is the mean, and σ\sigma is the standard deviation.

  • For the 30.5 kg bag:
    1.96=30.5μσ-1.96 = \frac{30.5 - \mu}{\sigma} This gives the equation: 30.5μ=1.96σ(Equation 1)30.5 - \mu = -1.96\sigma \quad \text{(Equation 1)} or μ=30.5+1.96σ(Equation 1’)\mu = 30.5 + 1.96\sigma \quad \text{(Equation 1')}

  • For the 46.3 kg bag:
    1.19=46.3μσ1.19 = \frac{46.3 - \mu}{\sigma} This gives the equation: 46.3μ=1.19σ(Equation 2)46.3 - \mu = 1.19\sigma \quad \text{(Equation 2)}

Step 3: Solve the system of equations

We now solve the two equations:

  1. μ=30.5+1.96σ\mu = 30.5 + 1.96\sigma
  2. 46.3μ=1.19σ46.3 - \mu = 1.19\sigma

Substitute Equation 1 into Equation 2: 46.3(30.5+1.96σ)=1.19σ46.3 - (30.5 + 1.96\sigma) = 1.19\sigma Simplify: 46.330.51.96σ=1.19σ46.3 - 30.5 - 1.96\sigma = 1.19\sigma 15.8=3.15σ15.8 = 3.15\sigma σ=15.83.155.02\sigma = \frac{15.8}{3.15} \approx 5.02

Step 4: Find μ\mu

Substitute σ=5.02\sigma = 5.02 into Equation 1: μ=30.5+1.96×5.0230.5+9.84=40.34\mu = 30.5 + 1.96 \times 5.02 \approx 30.5 + 9.84 = 40.34

Final Answer:

  • μ40.34kg\mu \approx 40.34 \, \text{kg}
  • σ5.02kg\sigma \approx 5.02 \, \text{kg}

Would you like further details or clarifications?


Follow-up Questions:

  1. What is the significance of the Z-score in the context of this problem?
  2. How do we interpret the mean and standard deviation in real-life scenarios?
  3. Can you explain why the Z-score table was used and its importance?
  4. What would happen if we changed the percentage from 97.5% to 99%?
  5. How would you solve this problem if only one percentage was given?

Tip:

Always ensure you know how to read and use the Z-table for solving normal distribution problems like this one.

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Math Problem Analysis

Mathematical Concepts

Normal Distribution
Z-scores
Statistics

Formulas

Z = (X - μ) / σ

Theorems

Normal Distribution Theorem
Z-score Interpretation

Suitable Grade Level

Grades 10-12